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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
13
votes
1
answer
3k
views
Does this metric have an official name? Lévy metric? Ky Fan metric?
Let $X$ and $Y$ be random variables taking values in a separable metric space $(S,d)$. The metric I have in mind is
$$\rho(X,Y) = \mathbb{E}[\min\{d(X,Y),1\}]$$
if $X$ and $Y$ take values in the a me …
5
votes
2
answers
639
views
Is the Hausdorff metric on sub-$\sigma$-fields separable?
Let $(X,\mu,\mathcal{F})$ be a probability space. The paper Equiconvergence of Martingales by Edward Boylan introduced a pseudometric on sub-$\sigma$-fields (sub-$\sigma$-algebras) of $\mathcal{F}$ a …
9
votes
2
answers
535
views
What mode of convergence is this?
I'm interested in a new (to me) mode of convergence which is stronger than convergence in measure/probability. I want to know if it has a name and if it is used much in the literature. I will write …
15
votes
2
answers
3k
views
Generalizations of the Tietze extension theorem (and Lusin's theorem)
I am reasking a year-old math.stackexchange.com question asked by someone else.
(For my needs every space $X$ and $Y$ will be Polish---that is a completely separably metrizable space.)
The Tietze ex …
8
votes
0
answers
382
views
What is the name for a Banach space property closed under ultraproducts?
In Banach space theory, a super-property is a property of a Banach space that is preserved under ultrapowers. (Update (2015-09-28): The property must also be closed under isometric embeddings.) (Sup …
4
votes
2
answers
505
views
Martingale-cotype vs cotype on super-reflexive spaces
I'm have difficultly nailing down the direction of some implications. For $2 \leq q < \infty$, there are (at least) two ways to say that a Banach space $B$ has "cotype $q$".
$B$ has cotype q.
$B$ is …
2
votes
0
answers
169
views
Classify spaces that make extension theorems hold
Recall a Polish space is a completely metrizable separable space.
Say a Polish space $Y$ is a terminal space if for any Polish space $X$ and any closed $C \subseteq X$, one can extend a continuous m …
4
votes
0
answers
309
views
Conditional expectation with respect to random closed sets
Short question
If $X$ is a random closed set, and $Y$ is an integrable random variable, I would like a definition of the "conditional expectation" $$\mathbf{E}[Y \mid X\ni x].$$ Has this been worked …
5
votes
2
answers
434
views
Sz.-Nagy dilation for uniformly convex Banach spaces
The Sz.-Nagy dilation theorem says that for a Hilbert space $H$ with nonexpansive operator $T$, there is a larger space $H'$ containing $H$ and a unitary operator $U$ on $H'$ such that for all $x \in …
3
votes
Accepted
Martingale-cotype vs cotype on super-reflexive spaces
I'll attempt to answer this with what I've found:
(1) The answer to my main question is that it is not true. In this document of Pisier's, he states
It is possible to find a uniformly convex s …
1
vote
1
answer
171
views
Extension of power bounded operators over a finite subspace
Suppose $Y$ is a Banach space and $X$ is a finite-dimensional subspace of $Y$. Further assume $T:X \rightarrow X$ is a linear operator which is power bounded from above and below, in other words ther …